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Transcript of Richard J. Terwilliger by Let’s look at some examples.
![Page 1: Richard J. Terwilliger by Let’s look at some examples.](https://reader036.fdocuments.us/reader036/viewer/2022062320/56649cc15503460f949886d8/html5/thumbnails/1.jpg)
Richard J. Terwilliger
by
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Let’s look at someexamples.
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We’ll break this GREEN vector up into different
RED vectors.
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Here the GREEN vector is broken down into
two RED vectors
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The two RED vectors are
called the of the
GREEN vector.
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Or, the of the two RED vectors is
the GREEN vector.
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The same GREEN vector can be broken up into
2 different RED vectors.
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Or two different vectors!
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There are
possibilities!
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A little later on we’ll show that two components at right
angles are very helpful!
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Here the same GREEN vector is broken up into 3 different
component vectors
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Again, the of the three RED vectors is the GREEN vector!
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And, the GREEN vector can be broken into RED vectors
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Let’s see anotherexample.
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This BLUE vector is broken down into several
ORANGE vectors.
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The tail of the BLUE vector starts at the same point as the tail
of the first ORANGE vector.
STARTING
POINT
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And they both end at the same point
ENDING
POINT
STARTING
POINT
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The of the ORANGE vectors is the
BLUE vector.
E
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E
Or again, theof the BLUE vector are the ORANGE vectors..
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E
So what have we learned so far?
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E
A vector can be broken down into any number of components.
The different arrangements of components are unlimited.
Components are one of two or more vectors
having a sum equal to a given vector.
A resultant is the sum of the component vectors.
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The process of finding the components of a vector given it’s magnitude and direction is called:
E
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E
is an easy way to find the resultant of several vectors.
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E
To show how, let’s go back to two at right angles..
90o
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E
We’ll sketch in the x-axis soit goes through the TAIL of our original vector.
X-axisTAIL
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EX-axis
Y-axis
TAIL
Then we’ll sketch the y-axis so it goes through the TAIL of our original vector, too.
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Y-axis
E
In the diagram the red vectoris called the horizontal or X- component.
X-axis
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Y-axis
E
The blue vector is called the vertical or Y- component.
X-axis
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Y-axis
E
To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.
X-axis
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Y-axis
E
To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.
X-axis
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Y-axis
E
To distinguish between the X and Y components, we label each vector with X and Y subscripts respectively.
X-axis
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E
Knowing the original vector’s magnitude and direction we can solve for Vx and Vy using trigonometry.
90o
0o180o
270o
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E
Let’s label the original vector with it’s magnitude and
direction.
0o180o
90o
270o
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90o
270o
E
Let’s label the original vector with it’s magnitude and
direction.
0o180o
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90o
270o
E
To determine the value of the X-component (Vx) we need
to use COSINE.
0o180o
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90o
270o
E
Remember COSINE?
0o180o
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E0o180o
Cosine = Adjacent Hypotenuse
90o
270o
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90o
270o
We need to rearrange the equation:
solving for the adjacent side.
.
Cosine = Adjacent Hypotenuse
E0o180o
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90o
270o
Cos = Adj Hyp
E0o180o
Adj = (Hyp) (Cos )therefore
Vx = V cos
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90o
270o
E0o180o
Adj = (Hyp) (Cos )therefore
Cos = Adj Hyp
Vx = V cos
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90o
270o
E0o180o
Vx = V cos Vx = 36 m/s (cos 42o)
Vx = 26.7 m/s
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90o
270o
E0o180o
We can now solve for Vy using Sine.
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90o
270o
E0o180o
Sine = Opposite Hypotenuse
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90o
270o
We are solving for the side opposite the 42 degree angle, Vy, therefore we’ll rearrange the equation solving for the opposite side.
E0o180o
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90o
270o
th
eref
ore
E0o180o
Vy = V sin
Opp = Hyp (Sin )
Sine = Opposite Hypotenuse
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90o
270o
th
eref
ore
E0o180o
Vy = V sin
Opp = Hyp (Sin )
Sine = Opposite Hypotenuse
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90o
270o
E0o180o
Vy = V sin
Vy = 36 m/s (sin 42o)
Vy = 24.1 m/s
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90o
270o
E0o180o
We’ve solved for the horizontal vertical and components
of our original vector!
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90o
270o
E0o180o
Let’s REVIEW what we did.
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V
REVIEW
A vector can be broken down into two vectors at right angles
A MATHEMATICAL METHOD
90o
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V
REVIEW
A MATHEMATICAL METHOD
The component that lies on the X-axis is called the horizontal or X - component
The X-component is found using Vx = Vcos
Vx = Vcos
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V
REVIEW
A MATHEMATICAL METHOD
The component that lies on the Y-axis is called the vertical or Y- component
The Y-component is found using Vy = Vsin
Vy = Vsin
Vx = Vcos
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Now that we know how to break vectors down into their X and Y components
we can solve for the resultant of many vectors using:
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Let’s look at a sample problem!
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travels 4.2 m at an angle of 37o north of east.
The first putt
10o east of north.
The second putt travels 3.9 m at
The third putt travels 2.6 m at an angle of 30o south of west.
The last putt heading 71o east of south travels 1.8 m before dropping into the
hole. What displacement was needed to sink the ball on the first putt?
A golfer, putting on a green, requires four shots to “hole the ball”.
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To solve the problem, using the component method, we’ll carry
out the following sequence: 1. List the vectors
2. Solve for the X and Y components of each vector
3. Sum the X components and sum the Y components
5. Solve for the direction of the resultant using tangent
4. Solve for the magnitude of the resultant using the Pythagorean theorem
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First, let’s list all of the vectors changing all of the directions to the number of degrees from east in a counterclockwise direction.
1. List
the ve
ctors
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travels 4.2 m at an angle of 37o north of east.
The first putt
10o east of north.
The second putt travels 3.9 m at
The third putt travels 2.6 m at an angle of 30o south of west.
The last putt heading 71o east of south travels 1.8 m before dropping into the
hole. What displacement was needed to sink the ball on the first putt?
A golfer, putting on a green, requires four shots to “hole the ball”.
90o
0o
180o
270o
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travels 4.2 m at an angle of 37o north of east.
The first putt
10o east of north.
The second putt travels 3.9 m at
The third putt travels 2.6 m at an angle of 30o south of west.
The last putt heading 71o east of south travels 1.8 m before dropping into the
hole. What displacement was needed to sink the ball on the first putt?
A golfer, putting on a green, requires four shots to “hole the ball”.
90o
0o
180o
270o
d1 = 4.2 m @ 37o
d2 = 3.9 m @ 80o
d3 = 2.6 m @ 210o
d4 = 1.8 m @ 341o
10o east of north 30
o south of west
71o east of south
37o north of east.4.2 m 3.9 m
2.6 m
1.8 m
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Next, to make things easier to see, let’s set up a table that lists our vectors..
1. List
the ve
ctors
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First label all of the columns
Vector(V)
X-component Solve Y-component Solve
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Remember the X component is cosine
Vector(V)
X-component Solve Y-component SolveVector(V)
X-component Solve Y-component Solve(Vcos)
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And the Y component is sine
Vector(V)
X-component Solve Y-component SolveVector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Next fill in the firstcolumn with the different vectors.
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Substitute into the equations for X and then Y
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
Remember to include units!
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Now solve for the value of each X and Y component.Again, remember to include UNITS!
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
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Sum the X components and then the Y components.
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Sum () x = 3.48 m Y = 4.49 m
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Sum () x = 3.48 m Y = 4.49 m
The four vectors are now broken down into two.
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Sum () x = 3.48 m Y = 4.49 m
One vector 3.48 m long lies on the X-axis.
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Sum () x = 3.48 m Y = 4.49 m
The other vector, 4.49 m long, lies on the Y-axis.
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Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
Vector(V)
X-component Solve Y-component Solve(Vcos) (Vsin )
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Sum () x = 3.48 m Y = 4.49 m
We’ll plot these two vectorsand determine their
resultant.
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Y
x = 3.48 m Y = 4.49 m
We’ll plot these two vectorsand determine their
resultant.
Xx = 3.48 m
Y =
4.4
9 m
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Y
Xx = 3.48 m
The resultant of these two vectors is the same as the resultant of our original four vectors!.
Y =
4.4
9 m
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Y
X
Y =
4.4
9 m
x = 3.48 m
The Pythagorean Theorem will be used to determine the
MAGNITUDE of the resultant.
c2 = a2 +b2
R2 = X2 + Y2
R = X2 + Y2
R = (3.48m)2 + (4.49m)2
R = 5.68 m
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Y
X
Y =
4.4
9 m
x = 3.48 m
We have now found the magnitude of our resultant
to be 5.68 m long.
c2 = a2 +b2
R2 = X2 + Y2
R = X2 + Y2
R = (3.48m)2 + (4.49m)2
R = 5.68 m
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Y
X
Y =
4.4
9 m
x = 3.48 m
The last step is to find the DIRECTION our resultant
is pointing.
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Y
X
Y =
4.4
9 m
Y = 3.48 m
To find the angle we’ll use tangent.
Tan =adjacent
opposite
adj
opp = Tan-1
X
Y = Tan-1
3.48 m
4.49 m = Tan-1
= 52.2o
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Y
X
Y =
4.4
9 m
Y = 3.48 m
The RESULTANT is5.68 m @ 52.2o.
Let’s go back to our original problem with
the results.
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travels 4.2 m at an angle of 37o north of east.
The first putt
10o east of north.
The second putt travels 3.9 m at
The third putt travels 2.6 m at an angle of 30o south of west.
The last putt heading 71o east of south travels 1.8 m before dropping into the
hole. What displacement was needed to sink the ball on the first putt?
A golfer, putting on a green, requires four shots to “hole the ball”.
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Let’s SUMMARIZE theCOMPONENT METHOD
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List all the vectors.
Break each vector down into an X and Y component
using Cosine for X and Sine for Y.
Sum all of the X components and then sum all of the Y components
Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant
Using Tangent solve for the DIRECTION of the resultant.
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Let’s go over that once again.
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Vector
(V)
X-componentSolve
Y-component Solve
x
Y
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
List all the vectors.
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Break each vector down into an X and Y component
using Cosine for X and Sine for Y.
Vector
(V)
X-componentSolve
Y-component Solve
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
(Vcos) (Vsin )
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
x
Y
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Sum all of the X components and then sum all of the Y components
Vector
(V)
X-componentSolve
Y-component Solve
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
(Vcos) (Vsin )
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
Sum () x = 3.48 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Y = 4.49 m
x
Y
Y = 3.48 m
Y =
4.4
9 m
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Vector
(V)
X-componentSolve
Y-component Solve
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
(Vcos) (Vsin )
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
Sum () x = 3.48 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Y = 4.49 m
Using the Pythagorean Theorem solve for the MAGNITUDE of the resultant
c2 = a2 +b2
R2 = X2 + Y2
R = X2 + Y2
R = (3.48m)2 + (4.49m)2
R = 5.68 m
x
Y
Y = 3.48 m
Y =
4.4
9 m
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Vector
(V)
X-componentSolve
Y-component Solve
4.2 m @ 37o
3.9 m @ 80o
2.6 m @ 210o
1.8 m @ 341o
(Vcos) (Vsin )
4.2 m cos 37o
3.9 m cos 80o
2.6 m cos 210o
1.8 m cos 341o
4.2 m sin 37o
3.9 m sin 80o
2.6 m sin 210o
1.8 m sin 341o
3.35 m
0.68 m
-2.25 m
1.70 m
Sum () x = 3.48 m
2.53 m
3.84 m
-1.30 m
-0.58 m
Y = 4.49 m
c2 = a2 +b2
R2 = X2 + Y2
R = X2 + Y2
R = (3.48m)2 + (4.49m)2
R = 5.68 m
x
Y
Y = 3.48 m
Y =
4.4
9 m
Using Tangent solve for the DIRECTION of the resultant.
Tan =adjacent
opposite
adj
opp = Tan-1
x
Y = Tan-1
3.48 m
4.49 m = Tan-1
= 52.2o
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